Question

A 0.34 kg projectile is fired into the air from the top of a 6.45 m cliff above a valley. Its Initial velocity is 11.4 m/s at 61? above the horizontal. How long is the projectile in the air for ?

How far from the bottom of the cliff does the projectile land?

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

Answer

Use the kinematic eqns to solve this

We have

y0 = 7.07

y = 0

vx0 = vo*cos(39) = 11.5*cos(39) = 8.937

vy0 =11.5*sin(39) = 7.237

The time to hit the ground comes from y = y0 + vy0*t - 1/2*g*t^2

so 1/2*g*t^2 - vy0*t + (y - y0) = 0 or 4.9*t^2 - 7.237*t - 7.07 = 0

1) t = 2.148 s

2) the horizontal distance = vx0*t = 8.937 * 2.148 = 19.196 m